If left(frac{2}{3}, 0right) is the centroid o | vedclass

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(B) The equation $4x^2-y^2=0$ can be factored as $(2x-y)(2x+y)=0$. This represents two lines: $L_1: 2x-y=0$ and $L_2: 2x+y=0$. The third line is $L_3:

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(B) The equation $4x^2-y^2=0$ can be factored as $(2x-y)(2x+y)=0$. This represents two lines: $L_1: 2x-y=0$ and $L_2: 2x+y=0$. The third line is $L_3: lx+my+n=0$. The vertices of the triangle are the intersection points of these lines. Intersection of $L_1$ and $L_2$ is $(0, 0)$. Intersection of $L_1$ and $L_3$: $y=2x$ and $lx+m(2x)+n=0 \implies x_1 = -n/(l+2m), y_1 = -2n/(l+2m)$. Intersection of $L_2$ and $L_3$: $y=-2x$ and $lx+m(-2x)+n=0 \implies x_2 = -n/(l-2m), y_2 = 2n/(l-2m)$. The centroid is $\left(\frac{x_1+x_2+0}{3}, \frac{y_1+y_2+0}{3}ight) = \left(\frac{2}{3}, 0ight)$. From the $y$-coordinate: $\frac{-2n/(l+2m) + 2n/(l-2m)}{3} = 0 \implies n/(l-2m) = n/(l+2m)$. Assuming $n 
eq 0$,we get $l-2m = l+2m \implies m=0$. From the $x$-coordinate: $\frac{-n/(l+2m) - n/(l-2m)}{3} = 2/3 \implies -n/l - n/l = 2 \implies -2n/l = 2 \implies l = -n$. Thus,$l+m+n = -n+0+n = 0$.

If $\left(\frac{2}{3}, 0\right)$ is the centroid of the triangle formed by the lines $4x^2-y^2=0$ and $lx+my+n=0$,then $l+m+n=$

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